Solving linear programming problems: the Simplex Method

FÖRELÄSNING 4(Chapter 4.8 - 4.10 )Solving linear programming problems: the Simplex Method4.6 Determining the starting feasible basic solution - The Two-Phase Method(Chapter4.9 in the book, p.113-117)Example 9 (Example in the Chapter 4.9 in the book, p.114)min z =2x 1 + x 2 + x 3s.t. 2x 1 − x 2 − x 3 ≥ 12x 1 + x 2 ≤ 5−x 2 + x 3 =1x 1 , x 2 , x 3 ≥ 0The standard form of the problem ismin z =2x 1 + x 2 + x 3s.t. 2x 1 − x 2 − x 3 − s 1 =12x 1 + x 2 + s 2 =5−x 2 + x 3 =1x 1 , x 2 , x 3 , s 1 , s 2 ≥ 0To find the starting feasible basic solution, we introduce artificial variables a 1 and a 3 to thefirst and the third constraints. The revised equations of the constraints become the following.2x 1 − x 2 − x 3 − s 1 + a 1 =12x 1 + x 2 + s 2 =5−x 2 + x 3 + a 3 =1x 1 , x 2 , x 3 , s 1 , s 2 , a 1 , a 3 ≥ 0Phase 1 :Minimizing the sum of the artificial variables with revised constraints.Phase 1 problem:min w = a 1 + a 3 (until a 1 = a 3 =0)s.t. 2x 1 − x 2 − x 3 − s 1 + a 1 =12x 1 + x 2 + s 2 =5−x 2 + x 3 + a 3 =1x 1 , x 2 , x 3 , s 1 , s 2 , a 1 , a 3 ≥ 01

Solving the Phase 1 problem will result in one of the following two cases:1. The optimal solution of the problem has value w =0, (a 1 = a 3 = 0), then it is a feasiblebasic solution for the original problem.2. The optimal solution of the problem has value w>0, the original problem has no feasiblesolution.Notice thatThe Phase 1 problem is always a minimization problem no matter the original problem isa maximization or minimization problem.Phase 2: Drop the artificial variables. Starting from the feasible basic solution obtained at theend of phase 1, use the simplex method to solve the original problem.Phase 2 problem:min z =2x 1 + x 2 + x 3 (original objective function)s.t. 2x 1 − x 2 − x 3 − s 1 =12x 1 + x 2 + s 2 =5−x 2 + x 3 =1x 1 , x 2 , x 3 , s 1 , s 2 ≥ 0The detail calculation of the simplex iterations for Example 9 can be found in the book,p.115-117.4.7 The simplex tableau in matrix form (Chapter 4.8 in the book, p.110-113)In general, the standard form of an LP-problem can be written in the following matrix form:max (or min)s.t.c T xAx = bx ≥ 0where x =(x 1 ,...,x n ) T , c =(c 1 ,...,c n ) T , b =(b 1 ,...,b m ) T , 0 =(0,...,0) Tan (m × n) matrix.and A =[a ij ]isLet x =(x B , x N ) T , where x B is the vector of basic variables, and x N is the vector of nonbasicvariables,A =(B | N), where matrix B consists of the columns of A corresponding to the basic variables,and matrix N consists of the columns of A corresponding to the nonbasic variables,c =(c B , c N ) T , where c B is the vector of the objective function coefficients for the correspondingbasic variables in x B , and c N is the vector of the objective function coefficients for thecorresponding nonbasic variables in x N .2

The initial simplex tableau in matrix form can be written as follows.Basvar z x T B x T N x T s¯bz 1 −c T B −c T N 0 T 0x s 0 B N I bwhere x s be the vector of the slack variables.At any iteration, the simplex tableau in matrix form becomesBasvar z x T B x T N x T s¯bz 1 0 T −(c T N − cT B B−1 N) c T B B−1 c T B B−1 bx B 0 I B −1 N B −1 B −1 bNotice thatIf x j is the slack variable in a ”≥” constraint, then we should add a negative sign for thecorresponding column in B −1 .Consider the following problem.Example 10max z =6x 1 + x 2 +4x 3 +5x 4s.t. 2x 1 + x 2 + x 3 + x 4 ≤ 20x 1 +2x 3 + x 4 ≤ 10x 1 + x 2 + x 3 ≤ 5x 1 , x 2 , x 3 , x 4 ≥ 0Let x 5 ,x 6 and x 7 denote the slack variables for the respective constraints. After the simplexmethod is applied, a portion of the final simplex tableau (optimal simplex tableau) is as follows.Basvar z x 1 x 2 x 3 x 4 x 5 x 6 x 7¯bz 1x 5 0 1 -1 -1x 4 0 0 1 -1x 1 0 0 0 1Identify the missing numbers in the simplex tableau, and indicate the optimal solution of theproblem.3